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Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)
\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy \(M=4\)
bạn cứ đặt công thức gốc là k sau đó thay vào các câu là được thui
*a/b=c/d=k=>a=bk;c=dk
Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3
Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3
=>2a+3b/2a-3b=2c+3d/2c-3d
*a/b=c/d=>a/c=b/d=k
=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)
k^2=a/c.b/d=ab/cd (2)
Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2
*a/b=c/d=>a/c=b/d=k=a+b/c+d
=>k^2=(a+b/c+d)^2
k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2
=>(a+b/c+d)^2=a^2+b^2/c^2+d^2
Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)
a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)
Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)
Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Mình hướng dẫn thôi nhé:
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm
Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:
\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a/ \(VT=\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1=\left(1\right)\)
\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b/ \(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
c/ \(VT=\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5d}=\dfrac{b\left(2k-5\right)}{d\left(2k-5\right)}=\dfrac{b}{d}\left(1\right)\)
\(VP=\dfrac{3a+4b}{3c+4d}=\dfrac{3bk+4b}{3dk+4d}=\dfrac{b\left(3k+4\right)}{d\left(3k+4\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2a-5b}{2c-5đ}=\dfrac{3a+4b}{3c+4d}\)
d/ \(VT=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-k^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\left(1\right)\)
\(VP=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)
Ta có :
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
\(a+b+c=\dfrac{a+2b-c}{c}=\dfrac{b+2c-a}{a}=\dfrac{c+2a-b}{b}=\)
\(=\dfrac{a+2b-c+b+2c-a+c+2a-b}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\dfrac{a+2b-c}{c}=2\Rightarrow a+2b=3c\)
\(\Rightarrow\dfrac{b+2c-a}{a}=2\Rightarrow b+2c=3a\)
\(\Rightarrow\dfrac{c+2a-b}{b}=2\Rightarrow c+2a=3b\)
\(P=\left(2+\dfrac{a}{b}\right)\left(2+\dfrac{b}{c}\right)\left(2+\dfrac{c}{a}\right)=\)
\(=\left(\dfrac{a+2b}{b}\right)\left(\dfrac{b+2c}{c}\right)\left(\dfrac{c+2a}{a}\right)=\dfrac{3c.3a.3b}{a.b.c}=27\)